Nature of Roots

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=3\), \(b=-6\), and \(c=1\), so \(D=(-6)^2-4(3)(1)=36-12=24\). Therefore, option A is correct. Remember that a positive discriminant indicates two distinct real roots; 36 is only the value of \(b^2\), not the complete discriminant.

The discriminant of ax² + bx + c = 0 is D = b² − 4ac. In this equation, a = 3, b = −6 and c = 1. Thus D = (−6)² − 4(3)(1) = 36 − 12 = 24. A positive discriminant shows that the two roots are real and distinct. To decide whether they are rational or irrational, observe that 24 is not a perfect square. Indeed, the quadratic formula gives x = [6 ± √24]/6 = 1 ± √6/3, and √6 is irrational. Hence both roots are real, irrational and distinct. Therefore option A is correct. Option B would require a positive perfect-square discriminant, while options C and D correspond to D = 0 and D < 0 respectively.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1, b=8, c=16\), so \(D=8^2-4(1)(16)=64-64=0\). Therefore, the equation has equal roots. In an exam, identify the signs of \(a,b,c\) before substitution; \(64\) is only the value of \(b^2\), not the complete discriminant.

Here, \(a=1, b=8, c=16\), so the discriminant is \(D=b^2-4ac=8^2-4(1)(16)=0\). Therefore, the roots are real and equal. In fact, \(x^2+8x+16=(x+4)^2\), so both roots are \(x=-4\). Exam tip: when \(D=0\), a quadratic equation has real and equal roots.

Comparing the equation with \(ax^2+bx+c=0\), we get \(a=2, b=7, c=3\). Thus, \(D=b^2-4ac=7^2-4(2)(3)=49-24=25\). The distractor 49 is only \(b^2\), without subtracting \(4ac\). Exam tip: Calculate the value and sign of \(D\) first to determine the nature of the roots.

Here, a = 2, b = 7 and c = 3. The discriminant is D = b² − 4ac = 7² − 4(2)(3) = 25. Since D > 0, the roots are real and distinct; since 25 is a perfect square, they are also rational. In fact, the roots are −1/2 and −3. Therefore, option A is correct. Exam tip: If D > 0 and D is a perfect square, the roots are real, rational and distinct.

Comparing the equation with the standard form \\(ax^2+bx+c=0\\), we get \\(a=7\\), \\(b=2\\), and \\(c=3\\). Thus, \\(D=b^2-4ac=(2)^2-4(7)(3)=4-84=-80\\). Therefore, -80 is correct. Since \\(D<0\\), the equation has no real roots. In exams, remember to use the coefficient of \\(x\\), including its sign, when calculating \\(b^2\\).

Here, \(a=7\), \(b=2\), and \(c=3\). The discriminant is \(\Delta=b^2-4ac=2^2-4(7)(3)=4-84=-80\). Since \(\Delta<0\), the equation has no real roots. Two real and equal roots occur only when \(\Delta=0\), so option B is incorrect. In an exam, first check the sign of the discriminant to determine the nature of the roots.

For equal real roots, the discriminant must be \(D=0\). Here, \(a=1, b=-2r, c=9\), so \(D=b^2-4ac=(-2r)^2-4(1)(9)=4r^2-36\). Thus, \(4r^2-36=0\), giving \(r^2=9\). Exam tip: For equal roots, immediately use the condition \(D=0\).

For real and equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=4\), \(b=s\), and \(c=9\), so \(s^2-4(4)(9)=0\). Therefore, \(s^2=144\). Exam tip: For equal roots of a quadratic equation, set the discriminant equal to zero.

Here (D=(-5)^2-4(1)(3)=13), and (13) is not a perfect square. So the roots are real, irrational and distinct.

A quadratic equation \(ax^2+bx+c=0\) has no real roots when its discriminant \(D=b^2-4ac\) is negative. Here, \(a=1, b=t, c=12\), so \(D=t^2-48\). Thus, \(t^2-48<0\), which gives \(t^2<48\). If \(t^2=48\), the equation has two equal real roots, so option B is incorrect. Exam tip: determine the nature of the roots by checking the sign of the discriminant first.

The roots of a quadratic equation are \\(\frac{-b\pm\sqrt{D}}{2a}\\). When \\(D>0\\), \\(\sqrt{D}\\) is a positive real number, so the plus and minus forms give two real and distinct roots. For comparison, \\(D=0\\) gives equal roots, while \\(D<0\\) gives no real roots. In an exam, first calculate or determine the sign of the discriminant \\(D=b^2-4ac\\).

Here (D=(-5)^2-4(1)(6)=1), so the roots are real and distinct. A perfect-square (D) gives rational roots.

Here, \(a=1, b=4, c=4\). Therefore, the discriminant is \(D=b^2-4ac=4^2-4(1)(4)=0\). When \(D=0\), a quadratic equation has two real and equal roots. In fact, \(x^2+4x+4=(x+2)^2\), so both roots are \(x=-2\). Exam tip: To determine the nature of roots, first calculate \(D=b^2-4ac\).

Here, \(a=2\), \(b=1\), and \(c=3\). Therefore, the discriminant is \(D=b^2-4ac=1^2-4(2)(3)=-23\). Since \(D<0\), the quadratic equation has no real roots. Options B and C are incorrect because they correspond to \(D=0\) and \(D>0\), respectively. Exam tip: To determine the nature of the roots, first check the sign of the discriminant.

For ax²+bx+c=0, the discriminant is D=b²−4ac. In x²−2x−3=0, a=1, b=−2 and c=−3. Hence D=(−2)²−4(1)(−3)=4+12=16. Since D>0, the equation has two real and distinct roots. Because 16 is also a perfect square, the roots are rational; in fact, they are x=[2±√16]/2=(2±4)/2, giving x=3 and x=−1. The question asks about their nature, so “two real and distinct” is the correct statement in option A. Option B would require D=0, option C would require D<0, and option D is misleading because there are two rational roots, not only one.

For a quadratic equation \(ax^2+bx+c=0\) to have equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=1\), \(b=-4\), and \(c=k\). Thus, \((-4)^2-4(1)(k)=0\), giving \(16-4k=0\) and hence \(k=4\). Therefore, option A is correct. Exam tip: for equal roots, immediately use \(D=0\).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=1\), \(b=k\), and \(c=9\), so \(D=k^2-36=0\). Thus, \(k^2=36\), giving \(k=6\) or \(k=-6\). Therefore, option A is correct. Exam tip: For equal roots of a quadratic equation, set \(D=0\) and remember to consider both positive and negative square roots.

For a quadratic equation \(ax^2+bx+c=0\) to have two real and distinct roots, its discriminant must satisfy \(\Delta=b^2-4ac>0\). Here, \(a=1, b=-6, c=k\), so \(\Delta=(-6)^2-4(1)(k)=36-4k\). Thus, \(36-4k>0\), which gives \(k<9\). When \(k=9\), the roots are equal, and when \(k>9\), the roots are not real. Exam tip: for distinct real roots, always require \(\Delta>0\).